$\frac{(1+{\sin }^{2}x){\cos }^{2}x}{{\cos }^{2}x-{\sin }^{2}x}$

$\frac{1-{\sin }^{4}x}{{\cos }^{2}x}$

$\frac{\left(\frac{1+{\sin }^{2}x}{{\cos }^{2}x-{\sin }^{2}x}\right)}{\cos x}$

$1$